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Why can't I solve this "easy" question?

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  • 07-03-2016 10:02pm
    #1
    Dughorm
    Closed Accounts Posts: 895 ✭✭✭


    If x+y=5 and x^y = 8 find x and y? :eek:

    It's lack of a method rather than the actual answer that bothers me.


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    #2
    ted1
    Registered Users Posts: 23,360 ✭✭✭✭ted1


    Dughorm wrote: »
    If x+y=5 and x^y = 8 find x and y? :eek:

    It's lack of a method rather than the actual answer that bothers me.

    Get x or y on its own. Y= 5-x
    Then replace x in the next one.
    X^(5-x)=8
    X^5-x^2=8
    X^3=8
    X=2
    Y=3


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    #3
    cruizer101
    Registered Users Posts: 3,205 ✭✭✭cruizer101


    How do you do this:
    X^(5-x)=8
    X^5-x^2=8

    Where does the two come from, is x^(5-x) not equal to (x^5)/(x^-x).

    I would of thought needed logs


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    #4
    GarIT
    Registered Users Posts: 8,671 ✭✭✭GarIT


    cruizer101 wrote: »
    How do you do this:
    X^(5-x)=8
    X^5-x^2=8

    Where does the two come from, is x^(5-x) not equal to (x^5)/(x^-x).

    I would of thought needed logs

    x^(5-x) = (x^5) - (x^2) it makes more sense when spread out but I'm thinking it should be (x^5) - (x^x) instead


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    #5
    Dughorm
    Closed Accounts Posts: 895 ✭✭✭Dughorm


    ted1 wrote: »
    Get x or y on its own. Y= 5-x
    Then replace x in the next one.
    X^(5-x)=8
    X^5-x^2=8
    X^3=8
    X=2
    Y=3

    So we're talking

    X^(5-x) = 8
    X^5-X^X = 8
    logX^5 - log X^X = log 8
    5logX - Xlog X = log 8

    Where does it go from here?


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    #6
    ZorbaTehZ
    Registered Users Posts: 2,149 ✭✭✭ZorbaTehZ


    Almost every post in this thread so far has a false statement(s) or deduction(s) in it! E.g. x^(5-x) is not equal to x^5-x^2, just try x=1 or x=2 to see why!

    Regarding the original question, the type of equations you posted are transcendental in general and will not have a closed form solution. In your case you have a solution in integers just by inspection.


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